Let $M$ be a spacetime (i.e., a differentiable manifold equipped with a Lorentzian metric) that admits a choice of time orientation.
For any region $R\subset M$, we define the causal complement of $R$ as $R' = M\setminus J(R)$, where $J(R) = J^+(R)\cup J^-(R)$ is the union of the causal future and causal past of $R$. We say that a set is causally complete if $R'' = R$ (i.e., if taking its causal complement twice returns the original region).
We can also define the causal hull of $R$ as $\text{ch}(R) = J^+(R)\cap J^-(R)$. We then say that a set is causally convex if $\text{ch}(R) = R$.
Question: Is causal completeness equivalent to causal convexity? In other words, is a set causally convex if and only if it is causally complete?
It is easy to see that $\text{ch}(R)\subseteq R''$. One way to prove this is by contradiction: in order for this to be false, there must be a point $p \in \text{ch}(R)$ and a point $q \in R'$ which can be connected to each other by a causal curve; but $p$ is itself connected to $R$ by a causal curve, so the existence of a causal curve between $p$ and $q$ would allow for the existence of a causal curve connecting $q$ and some point in $R$, which contradicts the assumption that $q \in R'$. Therefore, causal completeness implies causal convexity: since it is always true that $R\subseteq \text{ch}(R)$, and having $R'' = R$ also implies $\text{ch}(R)\subseteq R'' = R$, we are then led to $R=\text{ch}(R)$. I tried proving the converse statement with a similar strategy, but I couldn't finish the proof or find a counterexample. Is there something basic I'm not seeing here?